Show that the right-circular cone of the least curved surface and given volume has an altitude equal to √2 times the radius of the base.

OR

A window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 m, find the dimensions of the rectangle that will produce the largest area of the window.

Let, r be the radius of the right-circular cone, l be the slant height, and h be the altitude of a given right circular cone.

Given, that volume of cone (V = constant) and curved surface area of cone(C) = minimum

To prove: h = √2r

∵ curved surface area of cone = πrl

∴ C = πrl

∵ l can determined by using Pythagoras theorem.

∴ C = = πr√(r²+h²) …(1)

∵ Volume is constant.

So we can relate the height and radius with the help of V.

As we know that -

⇒ …(2)

Using equation 1 and 2, we can write –

As we need to minimise C,

If C is minimum C² will also be minimum, and we can also say that converse is true.

So, it is easy to minimise C² term

∴ squaring both sides, we get –

⇒

For S to be minimum,

∴ Differentiating w.r.t r we get –

⇒ …(3)

∴

⇒ 4π²r⁶ = 18V²

⇒

⇒

⇒ 2r⁶ = r⁴h²

⇒ h² = 2r²

⇒ h = √2r

Now, we need to check the sign of

∴ differentiating equation 3 again w.r.t r –

⇒

∴

⇒

∴ S is minimum, or curved surface area of a right circular cone is minimum for a given volume at h = √2r.

OR

Let x and y be the length and breadth of a rectangle, and an equilateral triangle is surmounted over it.

∴ the side of equilateral triangle = x

Given the perimeter of the window = 12 m

⇒ x + 2y + 2x = 12

⇒ 3x + 2y = 12

⇒ 2y = 12 – 3x

⇒ y = 6 – (3/2)x …(1)

Let A denotes the area of the window.

∴

⇒

⇒ …(2)

As we need to maximise A.

For A to be minimum,

∴ Differentiating equation 2 w.r.t x we get –

⇒ …(3)

⇒

⇒ 12 - 6x + √3x = 0

⇒

Putting the value of x in equation 1 we get-

∴ Differentiating equation 3 again, we get –

∴ A is maximum or area of the window is maximum for a given dimension of the rectangle –

and